5.61 Physical Chemistry
Exam III
11/29/12
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Chemistry
Chemistry - 5.61
Physical Chemistry
Exam III
(1) PRINT your name on the cover page.
(2) It is suggested that you READ THE ENTIRE EXAM before beginning work, since you may
be better prepared for some questions than others.
(3) ANSWER ALL QUESTIONS as completely as possible.
GOOD LUCK !!!!
Short Answer
(40 pts)________40_______________
Problem 2
(30 pts)________30_______________
Problem 3
(30 pts)________30_______________
TOTAL:
(100 pts)______100_______________
Name :________Key_________
TA:_________________
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5.61 Physical Chemistry
Exam III
11/29/12
Short Answer (40 points)
1) (7 Points) For a particular hydrogen orbital, ψ nlm , the average potential energy turns out to
be V (r) =-0.0625 a.u. What is V (r) for the analogous orbital in He+?
The potential energy scales as Z2. He+ has Z=2, so the average potential will be 0.0625*22=0.25 a.u.
2) (5 points) In performing spectroscopy of the Hydrogen atom, which pairs of orbitals below
would have allowed transitions? [Note: the Solid harmonics are related to the spherical
harmonics by: px∝Y11+Y1-1; py∝Y11-Y1-1; dxz∝Y21+Y2-1; dyz∝Y21-Y2-1; dxy∝Y22+Y2-2; dx2-y2 ∝Y22Y2-2]
5s
3px
3py
4dxy
4dz2
For polarized light, the selection rule will be Δl=±1 Δm=0. There is no selection rule on n.
Thus, none of the above would be allowed.
Alternatively, if the light is not polarized, the selection rule will be Δl=±1, Δm=±1,0. In this
case the allowed transitions will be
5s→3px
5s→3py
3px→4dxy
3px→4dz2
3py→4dxy
plus the reverse transitions. Full credit was given for either answer.
2
3py→4dz2
5.61 Physical Chemistry
Exam III
11/29/12
3) (6 points) Sketch a picture of the 5p radial function for a hydrogen atom, Rnl(r) . Note any
important features.
4) (4 points) For a particular atom, the spin part of the wavefunction is (α(1)β(2)+α(2)β(1))/√2.
Write down one possible form for the spatial wavefunction in this case.
Spin part is symmetric, so space part must be antisymmetric. Thus, one possibility would be
[1s(1)2s(2)-1s(2)2s(1)]/√2.
5) (4 points) What is a Slater Determinant and what is it used for?
A Slater determinant is an antisymmetric wavefunction built out of orbitals. It corresponds
to a unique stick diagram. It is used to create physically appropriate electronic
wavefunctions for atoms and molecules with many electrons. [Note: A perfect answer would
include the boldface terms in some reasonable fashion together with a bit about what the
slater determinant is used for. Partial credit for imperfect answers, of course.]
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5.61 Physical Chemistry
Exam III
11/29/12
6) (10 points) Consider the MO picture of bonding in Cr2. Ignore all the orbitals except for the
outermost d shell, so that there are 10 atomic orbitals available for making molecular orbitals
(dz2A, dz2B, dxzA, dxzB, dyzA, dyzB, dxyA, dxyB, dx2-y2A, dx2-y2B). Assume the molecule is oriented
along the z axis:
When you build the Hamiltonian matrix, H, for this molecule, it will be a 10x10 matrix.
However, as was the case for first row diatomics treated in class, the matrix will be block
diagonal - many of the matrix elements will be zero and only a few will be non-zero. In the
matrix below, shade in the blocks of the matrix you expect to be non-zero.
dzA2
dzA2
dzB2
d xzA
d xzB
d yzA
d yzB
d xyA
d xyB
d xA2 −y2
d xB2 −y2
dzB2
d xzA
d xzB
d yzA
d yzB
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
d xyA
d xyB
d xA2 −y2
d xB2 −y2
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
Note: these blocks correspond to σ, π and δ bonding orbitals respectively, moving from
the upper left to the lower right. There are also a few elements (dxzA-dyzA, dxzB-dyzB, dxyAdx2-y2A, dxyB-dx2-y2B) within these blocks that are zero, but no points were associated with
these terms, as the question specifically asks us to identify the blocks.
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5.61 Physical Chemistry
Exam III
11/29/12
7) (4 Points) What is the Born Oppenheimer Approximation?
The Born Oppenheimer approximation is the physical observation that, because the nuclei are
much heavier than the electrons, they move more slowly. As a result, we can think of the
electrons as moving in the field arising from the fixed nuclei. We can therefor separate the
electronic and nuclear parts of the wavefunction and retain a semi-classical picture of the
atomic positions.
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5.61 Physical Chemistry
Exam III
11/29/12
PROBLEM 2 (30 points)— Effective Nuclear Charge
Consider the Lithium atom within the independent particle model (IPM). You may find the
following two electron integrals useful (energies in eV; atomic unit of energy=27.2 eV):
J1s1s = 10.0 Z
J1s2s = 4.4 Z
J1s3s = 2.0 Z
J1s4s = 1.13 Z
K1s2s = 0.5 Z
K1s3s = 0.1 Z
K1s4s ≈ 0.0 Z
A) What is the ionization potential (IP) of Li in its ground state [1s22s1]?
(7 points) IP = E(1s2)-E(1s22s) = -E2s -2J1s2s+K1s2s=30.6 – 2*(13.2)+1.5=5.7 eV
B) Compute the IP of Li in the 1s23s1 excited state. Is the ionization potential larger or
smaller?
(5 points) IP = E(1s2)-E(1s23s) = -E3s -2J1s3s+K1s3s=13.6 – 2*(6)+0.3=1.9 eV
The IP is lower, as the excited electron is easier to remove.
C) Compute the IP of Li in the 1s24s1 excited state. Is the ionization potential larger or
smaller?
(5 points) IP = E(1s2)-E(1s24s) = -E4s -2J1s4s+K1s4s=7.65 – 2*(3.39)+0.0=0.87 eV
The IP is even lower, as the highly excited electron is even easier to remove.
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5.61 Physical Chemistry
Exam III
11/29/12
D) You may have been taught in earlier chemistry courses that the valence electrons in an
atom like Li see an effective nuclear charge, Zeff, that is lower than the actual nuclear
charge. Use your results from parts A)-C) to make this connection quantitative. Assume
that the ionization potentials you computed arose exactly from a Hydrogenic atom with
nuclear charge Zeff
me 4 Z eff2
IP = E(cation) − E(neutral) =
32π 2ε 02 2 n 2
Compute the correct Zeff values for the 2s, 3s and 4s orbitals in Li. How do you rationalize
your results?
(13 points) For 2s: IP = 5.7 eV = Zeff2/(2 n2) a.u. = Zeff2/(2 22) a.u. = Zeff2/8 a.u. = 3.4*Zeff2
eV.
→Zeff=1.38
For 3s: IP = 1.9 eV = Zeff2/(2 n2) a.u. = Zeff2/(2 32) a.u. = Zeff2/18 a.u. = 1.51*Zeff2 eV.
→Zeff=1.12
For 4s: IP = 0.87 eV = Zeff2/(2 n2) a.u. = Zeff2/(2 42) a.u. = Zeff2/32 a.u. = 0.85*Zeff2 eV.
→Zeff=1.01
As we go up in n, Zeff goes down progressively, as the electron spends more and more of
its time outside the 1s shell, where the effective charge approaches Zeff=1. Clearly, the
effective charge is asymptotically approaching Zeff=1 as n increases. This agrees with
my chemical intuition that Li should have an effective charge much closer to 1 than 3.
However, it is also clear that this is only qualitatively correct.
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5.61 Physical Chemistry
Exam III
11/29/12
PROBLEM 3 (30 points)— Molecular Orbital Theory
Consider the σ bond in LiH in the valence approximation, so that the molecular orbitals can be
written
ψ el = c1 2sLi + c21sH
At a particular bond length, R,
ε Li = ∫ 2sLi Ĥ 2sLi dτ = −7.0 eV
ε H = ∫ 1sH Ĥ 1sH dτ = −10.0 eV
V= ∫ 2sLi Ĥ1sH dτ = −2.0 eV
S = ∫ 2sLi1sH dτ ≈ 0
1
2
2
1
2sLi +
1sH and ψ =
2sLi −
1sH are energy eigenstates for LiH.
5
5
5
5
What are the corresponding eigenvalues? [Note: You do not need to solve for the
eigenvectors. Just show that the given eigenvectors are correct.]
A) Show that ψ =
(12 points) The Hamiltonian is
⎛ ε Li V ⎞
H=⎜
⎟
⎝ V εH ⎠
Plugging in the two eigenvectors, we find
⎛ ε Li V ⎞ ⎛ 15 ⎞ ⎛ −7 −2 ⎞ ⎛ 15 ⎞ ⎛
Hc = ⎜
⎟⎜ 2 ⎟ =⎜
⎟⎠ ⎜⎜ 2 ⎟⎟ = ⎜⎜
−2
−10
V
ε
⎜
⎟
⎝
H ⎠⎝
⎝
5 ⎠
⎝ 5 ⎠ ⎝
−7
5
−
4
5
−2
5
−
20
5
⎞
⎛
⎟ = −11⎜
⎟⎠
⎜⎝
1
5
2
5
⎞
⎟
⎟⎠
E = −11 eV
⎛ ε Li
Hc = ⎜
⎝ V
V ⎞⎛
⎟⎜
ε H ⎠ ⎝⎜
2
5
−1
5
⎞ ⎛
⎛
−7 −2 ⎞
⎟ =⎜
⎜
⎟⎠ ⎝ −2 −10 ⎠⎟ ⎜⎝
E = −6 eV
8
2
5
−1
5
⎞ ⎛
⎟ =⎜
⎟⎠ ⎜⎝
−14
5
+
2
5
−4
5
+
10
5
⎞
⎛
⎟ = −6 ⎜
⎟⎠
⎜⎝
2
5
−1
5
⎞
⎟
⎠⎟
5.61 Physical Chemistry
Exam III
11/29/12
B) Using the orbitals above, what is the partial charge on Li, qLi, when LiH is in its electronic
ground state?
(6 points) The first vector is the ground state, which will have two valence electrons. The
number probability that each of those electrons ends up on Li is just |c1|2. Thus, the number
of electrons on Li is 2 |c1|2 and the charge is going to be 1-2 |c1|2. Plugging in:
qLi=1-2 |c1|2 = 1-2*(1/√5)2=3/5 = +0.6.
This seems about right: The Hydrogen is more electronegative and so it should grab more of
the bonding electrons.
C) If the nuclear charge on Li were increased (say to +3.1 e), how would each of the MO
quantities below change? You may briefly justify your answers. Recall that a negative
number becoming more positive corresponds to an increase, even though the absolute
value decreases.
εLi
Increase
Decrease
No Change
εH
Increase
Decrease
No Change
V
Increase
Decrease
No Change
qLi
Increase
Decrease
No Change
(6 points) Intended Answer: Increased charge will lower energy on Li. Chare on Li will
increase because the nuclear charge has increased (more electrons will flow to it, but
less than the change in nuclear charge). All other quantities do not change. A case could
be made for the following variations:
εH decreases slightly because of attraction of H 1s to Li nucleus.
V decreases slightly because Li 2s orbital contracts.
qLi decreases because you forgot that the nuclear charge went up.
Points given for variant answers if they were justified.
D) If we increased the bond length, R, how would each of the MO quantities below change?
You may briefly justify your answers. Recall that a negative number becoming more
positive corresponds to an increase, even though the absolute value decreases.
εLi
Increase
Decrease
No Change
εH
Increase
Decrease
No Change
V
Increase
Decrease
No Change
qLi
Increase
Decrease
No Change
(6 points) Intended Answer: Increased bond distance reduces the magntiude of the
resonance integral (increases in absolute value), which means less equal electron
sharing (less covalency) and thus more ionic character, which will increase the charge on
Li. All other quantities do not change. A case could be made that the energies will both
go up slightly due to lower electron-nuclear attraction, and points were given for this
answer if a justification was given.
9
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